MAYBE
* Step 1: UnsatPaths MAYBE
    + Considered Problem:
        Rules:
          0. f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F)      True                   (1,1)
          1. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [H >= 1 && A >= 1]     (?,1)
          2. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [0 >= 1 + H && A >= 1] (?,1)
          3. f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H) [A >= 1 && C >= 3]     (?,1)
          4. f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F)      [0 >= A]               (?,1)
        Signature:
          {(f0,6);(f1,6);(f2,6)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3,4},4->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(3,4)]
* Step 2: TrivialSCCs MAYBE
    + Considered Problem:
        Rules:
          0. f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F)      True                   (1,1)
          1. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [H >= 1 && A >= 1]     (?,1)
          2. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [0 >= 1 + H && A >= 1] (?,1)
          3. f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H) [A >= 1 && C >= 3]     (?,1)
          4. f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F)      [0 >= A]               (?,1)
        Signature:
          {(f0,6);(f1,6);(f2,6)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3},4->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 3: PolyRank MAYBE
    + Considered Problem:
        Rules:
          0. f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F)      True                   (1,1)
          1. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [H >= 1 && A >= 1]     (?,1)
          2. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [0 >= 1 + H && A >= 1] (?,1)
          3. f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H) [A >= 1 && C >= 3]     (?,1)
          4. f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F)      [0 >= A]               (1,1)
        Signature:
          {(f0,6);(f1,6);(f2,6)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3},4->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
          p(f0) = x1
          p(f1) = x1
          p(f2) = x1
        
        Following rules are strictly oriented:
        [0 >= 1 + H && A >= 1] ==>                     
               f0(A,B,C,D,E,F)   = A                   
                                 > -1 + A              
                                 = f0(-1 + A,B,I,G,H,F)
        
        
        Following rules are weakly oriented:
                      True ==>                     
           f1(A,B,C,D,E,F)   = A                   
                            >= A                   
                             = f0(A,B,C,D,E,F)     
        
        [H >= 1 && A >= 1] ==>                     
           f0(A,B,C,D,E,F)   = A                   
                            >= -1 + A              
                             = f0(-1 + A,B,I,G,H,F)
        
        [A >= 1 && C >= 3] ==>                     
           f0(A,B,C,D,E,F)   = A                   
                            >= A                   
                             = f0(A,B,-1 + C,G,0,H)
        
                  [0 >= A] ==>                     
           f0(A,B,C,D,E,F)   = A                   
                            >= A                   
                             = f2(A,G,C,D,E,F)     
        
        
* Step 4: PolyRank MAYBE
    + Considered Problem:
        Rules:
          0. f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F)      True                   (1,1)
          1. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [H >= 1 && A >= 1]     (?,1)
          2. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [0 >= 1 + H && A >= 1] (A,1)
          3. f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H) [A >= 1 && C >= 3]     (?,1)
          4. f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F)      [0 >= A]               (1,1)
        Signature:
          {(f0,6);(f1,6);(f2,6)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3},4->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
          p(f0) = x1
          p(f1) = x1
          p(f2) = x1
        
        Following rules are strictly oriented:
            [H >= 1 && A >= 1] ==>                     
               f0(A,B,C,D,E,F)   = A                   
                                 > -1 + A              
                                 = f0(-1 + A,B,I,G,H,F)
        
        [0 >= 1 + H && A >= 1] ==>                     
               f0(A,B,C,D,E,F)   = A                   
                                 > -1 + A              
                                 = f0(-1 + A,B,I,G,H,F)
        
        
        Following rules are weakly oriented:
                      True ==>                     
           f1(A,B,C,D,E,F)   = A                   
                            >= A                   
                             = f0(A,B,C,D,E,F)     
        
        [A >= 1 && C >= 3] ==>                     
           f0(A,B,C,D,E,F)   = A                   
                            >= A                   
                             = f0(A,B,-1 + C,G,0,H)
        
                  [0 >= A] ==>                     
           f0(A,B,C,D,E,F)   = A                   
                            >= A                   
                             = f2(A,G,C,D,E,F)     
        
        
* Step 5: Failure MAYBE
  + Considered Problem:
      Rules:
        0. f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F)      True                   (1,1)
        1. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [H >= 1 && A >= 1]     (A,1)
        2. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [0 >= 1 + H && A >= 1] (A,1)
        3. f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H) [A >= 1 && C >= 3]     (?,1)
        4. f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F)      [0 >= A]               (1,1)
      Signature:
        {(f0,6);(f1,6);(f2,6)}
      Flow Graph:
        [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3},4->{}]
      
  + Applied Processor:
      Failing "Open problems left."
  + Details:
      Open problems left.
MAYBE